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I'm looking for a reference or answer for the following question:

Let $M$ be an (compact and orientable, if it helps) smooth manifold and $\nu$ and $\mu$ two differential forms. I'm looking for conditions that allow me to find a vector field $X$ on $M$ such that $\nu=\mathcal{L}_X\mu$, where $\mathcal{L}$ denotes the Lie derivative on differential forms.

A special case in which I'm particularly interested in is if $\mu$ is a positive volume form with $\int_M\mu=1$ and $\nu$ is a volume form with $\int_M\nu=0$. Then the above question can be rephrased as: given a smooth function $\varphi$ with $\int_M\varphi\mu=0$, can we find a vector field $X$ such that $\varphi=\text{div}_{\mu}X$?

I've read somewhere that under certain assumptions one can always find such a vector field, but I cannot find that statement anymore and so I would appreciate if someone knew a short answer or a reference for this.

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    $\begingroup$ In the general case, for given $\mu$ and $\nu$, the equation $\mathcal{L}_X\mu=\nu$ is highly overdetermined, and there will be no solution. For example, if $\mu$ is closed and $\nu$ is not, then there is no solution, since $\mathcal{L}_X$ commutes with the exterior derivative $\mathrm{d}$. $\endgroup$
    – Robert Bryant
    Commented Oct 20, 2018 at 11:56
  • $\begingroup$ @Robert: Yes, I'm not very optimistic. Another way of looking at this question is that it is an infinitesimal version of the question, given two $k$-forms, can we find a diffeomorphism that pulls back one to the other one? So we could ask: is there a class of manifolds for which there is a nice characterization of the orbits of Diff$(M)$ acting on $\Omega^k(M)$? The infinitesimal question would be phrased in terms of the Lie algebra of Diff$(M)$. $\endgroup$
    – S.Surace
    Commented Oct 20, 2018 at 13:19

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If $\mu$ is a volume form and $\nu$ is a top degree form, then there exists a vector field $X$ such that $L_X\mu=\nu$ if and only if $\nu$ is exact.

You can always fix a metric $g$ on $M$ such that $\mu$ is the volume form determined by the metric and the orientation, $\mu=dVol_g$.

Denote by $\DeclareMathOperator{\Vect}{Vect}$ $\Vect(M)$ the space of smooth vector fields on $M$ and by $\ast$ the Hodge star operator determined by $g$ and the orientation. Then $\mu=\ast 1$.

For any vector field $X$ on $M$ denote by $\omega_X\in \Omega^1(M)$ the $1$-form dual to $X$, i.e., $$ \omega_X(Y)=g(X,Y),\;\;\forall Y\in \Vect(M). $$ Then (see Proposition 4.1.48 of these notes) $\DeclareMathOperator{\dive}{div}$ $$ \dive_\mu X=\dive_g X=\ast d\ast \omega_X\Longleftrightarrow \nu=\ast \dive X=d\ast\omega_X. $$ Thus if there exists a vector field $X$ such that $L_X\mu=\nu$ then $\nu$ is exact $\nu=d\ast\omega_X$, i.e., $\nu$ is exact. Conversely, if $\nu$ is exact, $\nu=d\beta$, then the vector field $X$ uniquely characterized by the equality $\ast\omega_X=\beta$ satisfies $L_X\mu=\nu$.

Remark. If $M$ is compact and connected then $\nu$ is exact if ad only if $\int_M\nu=0$. If $M$ is noncompact and connected then $\nu$ is automatically exact since $H_{DR}^{\dim M}(M)=0$. If $M$ has several connected components solve this problem on each component separately.

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  • $\begingroup$ After pondering your answer for a bit, isn't exactness of $\nu$ already sufficient for the case $k=n$ (regardless of compactness of $M$)? I may be missing something. But if $\nu=d\eta$, can't we can just chose a Riemannian metric as you do and then choose $X$ as one of $\pm\sharp\ast\eta$? $\endgroup$
    – S.Surace
    Commented Oct 19, 2018 at 11:07
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    $\begingroup$ On a compact, connecetd oriented manifold a top degree form is exact if and only if its integral is zero. $\endgroup$
    – Liviu Nicolaescu
    Commented Oct 19, 2018 at 11:45
  • $\begingroup$ I do not understand the meaning of $\sharp$ $\endgroup$
    – Liviu Nicolaescu
    Commented Oct 19, 2018 at 11:47
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    $\begingroup$ On noncompact (orientable) manifolds the top cohomology is trivial. Also, due to noncompactness not every top degree form is integrable over the manifold. On compact disconnected manifold, say with two components, you can have a form that integrates to 1 on one component and to -1 on the other component so the total integral is zero without being exact. The dimension of the top DeRham cohomology is equal to the number of components. $\endgroup$
    – Liviu Nicolaescu
    Commented Oct 19, 2018 at 17:03
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    $\begingroup$ In the noncompact (orientable)case every top degree form $\alpha$ is exact ,$\alpha =d\beta$ so $\\ast \alpha =\pm \ast d(\ast \ast\beta)=\pm d^*(\ast\beta)$. The result is true inthis case as well. $\endgroup$
    – Liviu Nicolaescu
    Commented Oct 20, 2018 at 0:51

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